414
10 Heat and Matter
Energy released in coalescing n droplets each of radius r into a large drop of
radius R
W = 4πr
2
(n − n
2/3
)S
(10.36)
W = 4π R
2
(n
1
3 − 1)S
(10.37)
Capillary rise:
2S cos θ = (h + r/3)rρg
(10.38)
where θ is the angle of contact, r is the radius of the bore and h is the height of the
liquid column.
10.2 Problems
10.2.1 Kinetic Theory of Gases
10.1 Define the mean free path of a gas both mathematically and in words. Calculate the mean free path of a molecule in a gas if the number of collisions is
2 × 10 10 /s and the mean molecular velocity is 1000 m/s.
[University of Manchester 2008]
10.2 (a) Consider a gas that has a molecular weight of 28 and a temperature of
27 ◦ C. What is the rms speed of molecule of the gas if it has a Maxwellian
velocity distribution? The ideal gas constant is 8.31 J/mol K.
(b) What is the mean free path of a molecule if the pressure is 2 atm (1 atm =
101.3 kPa), the temperature is 27 ◦ C, and the cross-section is 0.43 nm 2 ?
Using the average velocity from part (a) calculate the collision frequency
for the molecule. Boltzmann’s constant is 1.38 × 10 −23 J/K.
10.3 (a) By considering a volume, V , of ideal gas, containing N spherical
molecules with radius r , show that the mean free path of the molecules
can be defined by the equation
λ =
V
4π
√
2r 2 N
(b) Hence, or otherwise, calculate the mean free path of air at 100 ◦ C and
1.01 × 10 5 Pa. Assume that the molecules of air are spheres of radius
r = 2.0 × 10 −10 m.
(k = 1.38 × 10 −23 J/K)
[University of Aberystwyth, Wales]
10.4 Figure 10.1 shows the Maxwell–Boltzmann velocity distribution functions
of a gas for two different temperatures, of which first (curve f 1 ) is for
T 1 = 300 K.
10 Heat and Matter
Energy released in coalescing n droplets each of radius r into a large drop of
radius R
W = 4πr
2
(n − n
2/3
)S
(10.36)
W = 4π R
2
(n
1
3 − 1)S
(10.37)
Capillary rise:
2S cos θ = (h + r/3)rρg
(10.38)
where θ is the angle of contact, r is the radius of the bore and h is the height of the
liquid column.
10.2 Problems
10.2.1 Kinetic Theory of Gases
10.1 Define the mean free path of a gas both mathematically and in words. Calculate the mean free path of a molecule in a gas if the number of collisions is
2 × 10 10 /s and the mean molecular velocity is 1000 m/s.
[University of Manchester 2008]
10.2 (a) Consider a gas that has a molecular weight of 28 and a temperature of
27 ◦ C. What is the rms speed of molecule of the gas if it has a Maxwellian
velocity distribution? The ideal gas constant is 8.31 J/mol K.
(b) What is the mean free path of a molecule if the pressure is 2 atm (1 atm =
101.3 kPa), the temperature is 27 ◦ C, and the cross-section is 0.43 nm 2 ?
Using the average velocity from part (a) calculate the collision frequency
for the molecule. Boltzmann’s constant is 1.38 × 10 −23 J/K.
10.3 (a) By considering a volume, V , of ideal gas, containing N spherical
molecules with radius r , show that the mean free path of the molecules
can be defined by the equation
λ =
V
4π
√
2r 2 N
(b) Hence, or otherwise, calculate the mean free path of air at 100 ◦ C and
1.01 × 10 5 Pa. Assume that the molecules of air are spheres of radius
r = 2.0 × 10 −10 m.
(k = 1.38 × 10 −23 J/K)
[University of Aberystwyth, Wales]
10.4 Figure 10.1 shows the Maxwell–Boltzmann velocity distribution functions
of a gas for two different temperatures, of which first (curve f 1 ) is for
T 1 = 300 K.